 Hi and welcome to the session. I am Asha and I am going to help you with the following question that says find the sum to n terms of the series whose nth term is given by 2n-1 whole square. So let's now start with the solution and the nth term of the series is equal to 2n-1 whole square. Therefore, kth term of the series will be equal to 2k-1 whole square. So let the kth term be denoted by ak, so ak is equal to 2k-1 whole square or ak is equal to 4k-4k-1. Now since we have to find the sum to n terms, therefore taking summation on both the sides, summation ak, k running from 1 to n is equal to summation, k running from 1 to n, 4k-4k-1. This is further equal to 4 times summation k is equal to 1 to n k square minus 4 times summation k is equal to 1 to n k plus summation k is equal to 1 to n 1. So this is equal to 4 times summation of k square k running from 1 to n is n into n plus 1 into 2n plus 1 upon 6 minus 4n to summation kk running from 1 to n is n into n plus 1 upon 2 plus n, so 2-3 is a 6, 2-2 is a 4, 2-2 is a 4, so now taking n upon 3 common, here we have 2n plus 2 into 2n plus 1 minus 6 times of n plus 1 plus 3 is equal to n upon 3 into 4n square plus 2n plus 4n plus 2 minus 6n minus 6 plus 3, so plus 4n plus 2n is plus 6n which cancels out with minus 6n and we further have n upon 3 into 4n square. Now 3 plus 2 is 5, 5 minus 6 is minus 1, so this is further equal to upon 3 2n square minus 1 square which is in the form of a square minus b square and its formula is a minus b into a plus b, so we have 5 upon 3 2n minus 1 into 2n plus 1 sum to n term, so this n term is given as n upon 3 into 2n minus 1 into 2n plus 1. So this completes the session, take care and have a good day.